∫x^2(x^2+1)(x^2+4)dx =∫4x^2(4x^2+4)(x^2+4)dx =4∫x^2(4x^2+4)(x^2+4)dx =43∫(4x^2+4) (x^2+4)(4x^2+4)(x^2+4)dx =43∫(4x^2+4)dx(4x^2+4)(x^2+4 ...

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Second Fundamental Theorem of Calculus

Find ∫x2x2+...

Question

Find $\int \frac{x^{2}}{(x^{2} + 1) (x^{2} + 4)} d x$

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Solution

$\int \frac{x^{2}}{(x^{2} + 1) (x^{2} + 4)} d x$

$= \int \frac{4 x^{2}}{(4 x^{2} + 4) (x^{2} + 4)} d x$

$= 4 \int \frac{x^{2}}{(4 x^{2} + 4) (x^{2} + 4)} d x$

$= \frac{4}{3} \int \frac{(4 x^{2} + 4) - (x^{2} + 4)}{(4 x^{2} + 4) (x^{2} + 4)} d x$

$= \frac{4}{3} \int \frac{(4 x^{2} + 4) d x}{(4 x^{2} + 4) (x^{2} + 4)} - \frac{4}{3} \int \frac{(x^{2} + 4) d x}{(4 x^{2} + 4) (x^{2} + 4)}$

$= \frac{4}{3} \int \frac{d x}{(x^{2} + 4)} - \frac{4}{3} \int \frac{d x}{(4 x^{2} + 4)}$

$= \frac{4}{3} \int \frac{d x}{x^{2} + 4} - \frac{4}{12} \int \frac{d x}{x^{2} + 1}$

$= \frac{4}{3} \times \frac{1}{2} {tan}^{- 1} \frac{x}{2} - \frac{1}{3} {tan}^{- 1} x + c$

$= \frac{2}{3} {tan}^{- 1} \frac{x}{2} - \frac{1}{3} {tan}^{- 1} x + c$

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